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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials

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Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 3 - Factorisation of Polynomials.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials Page/Excercise MCQ

Solution 1

Correct option: (c)

Let p(x) = 2x2 + kx

Since (x + 1) is a factor of p(x),

P(-1) = 0

2(-1)2 + k(-1) = 0

2 - k = 0

k = 2

Solution 2

Correct option: (d)

(249)2 - (248)2

= (249 + 248)(249 - 248)

= 497 × 1

= 497

Solution 3

  

Solution 4

Solution 5

Correct option: (c)

  

Solution 6

  

Solution 7

Correct option: (d)

(x + y)3 - (x3 + y3)

= (x + y)3 - [(x + y)(x2 - xy + y2)]

= (x + y)[(x + y)2 - (x2 - xy + y2)]

= (x + y)[x2 + y2 + 2xy - x2 + xy - y2]

= (x + y)(3xy) 

Solution 8

Correct option: (d)

(25x2 - 1) + (1 + 5x)2

= [(5x)2 - (1)2] + (1 + 5x)2

= (5x - 1)(5x + 1) + (1 + 5x)2

= (1 + 5x)[(5x - 1) + (1 + 5x)]

= (1 + 5x)(5x - 1 + 1 + 5x)

= (1 + 5x)(10x)

Solution 9

Correct option: (b)

Since (x + 5) is a factor of p(x) = x3 - 20x + 5k,

p(-5) = 0

(-5)3 - 20(-5) + 5k = 0

-125 + 100 + 5k = 0

-25 + 5k = 0

5k = 25

k = 5 

Solution 10

  

Solution 11

  

Solution 12

  

Solution 13

  

Solution 14

  

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

  

Solution 22

Correct option: (d)

  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials Page/Excercise 3E

Solution 1(iii)

  

Solution 1(i)

(3x + 2)3

= (3x)3 + (2)3 + 3 3x 2 (3x + 2)

= 27x3 + 8 + 18x (3x + 2)

= 27x3 + 8 + 54x2 + 36x.

Solution 1(ii)

Solution 2(i)

(5a - 3b)3

= (5a)3 - (3b)3 - 3(5a)(3b)(5a - 3b)

= 125a3 - 27b3 - 45ab(5a - 3b)

= 125a3 - 27b3 - 225a2b + 135ab2 

Solution 2(ii)

  

Solution 2(iii)

  

Solution 3

8a3 + 27b3 + 36a2b + 54ab2

= (2a)3 + (3b)3 + 3 × (2a)2 × (3b) + 3 × (2a) × (3b)2

= (2a + 3b)3

= (2a + 3b)(2a + 3b)(2a + 3b) 

Solution 4

64a3 - 27b3 - 144a2b + 108ab2

= (4a)3 - (3b)3 - 3 × (4a)2 × (3b) + 3 × (4a) × (3b)2

= (4a - 3b)3

= (4a - 3b)(4a - 3b)(4a - 3b) 

Solution 5

  

Solution 6

125x3 - 27y3 - 225x2y + 135xy2

= (5x)3 - (3y)3 - 3 × (5x)2 × (3y) + 3 × (5x) × (3y)2

= (5x - 3y)3

= (5x - 3y)(5x - 3y)(5x - 3y)  

Solution 7

a3x3 - 3a2bx2 + 3ab2x - b3

= (ax)3 - 3 × (ax)2 × b + 3 × (ax) × (b)2 - (b)3

= (ax - b)3

= (ax - b)(ax - b)(ax - b)  

Solution 8

  

Solution 9

a3 - 12a(a - 4) - 64

= a3 - 3 × a × 4(a - 4) - (4)3

= (a - 4)3

= (a - 4)(a - 4)(a - 4) 

Solution 10(i)

(103)3

= (100 + 3)3

= (100)3 + (3)3 + 3 × 100 × 3 × (100 + 3)

= 1000000 + 27 + 900(103)

= 1000000 + 27 + 92700

= 1092727 

Solution 10(ii)

(99)3

= (100 - 1)3

= (100)3 - (1)3 - 3 × 100 × 1 × (100 - 1)

= 1000000 - 1 - 300(99)

= 1000000 - 1 - 29700

= 970299  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials Page/Excercise 3C

Solution 1

x2 + 11x + 30

= x2 + 6x + 5x + 30

= x (x + 6) + 5 (x + 6)

= (x + 6) (x + 5).

Solution 2

x2 + 18x + 32

= x2 + 16x + 2x + 32

= x (x + 16) + 2 (x + 16)

= (x + 16) (x + 2).

Solution 3

x2 + 20x - 69

= x2 + 23x - 3x - 69

= x(x + 23) - 3(x + 23)

= (x + 23)(x - 3) 

Solution 4

x2 + 19x - 150

= x2 + 25x - 6x - 150

= x(x + 25) - 6(x + 25)

= (x + 25)(x - 6) 

Solution 5

x2 + 7x - 98

= x2 + 14x - 7x - 98

= x(x + 14) - 7(x + 14)

= (x + 14)(x - 7) 

Solution 6

  

Solution 7

x2 - 21x + 90

= x2 - 6x - 15x + 90

= x(x - 6) - 15(x - 6)

= (x - 6)(x - 15) 

Solution 8

x2 - 22x + 120

= x2 - 10x - 12x + 120

= x(x - 10) - 12(x - 10)

= (x - 10)(x - 12) 

Solution 9

x2 - 4x + 3

= x2 - 3x - x + 3

= x(x - 3) - 1(x - 3)

= (x - 3)(x - 1) 

Solution 10

  

Solution 11

  

Solution 12

  

Solution 13

  

Solution 14

x2 - 24x - 180

= x2 - 30x + 6x - 180

= x(x - 30) + 6(x - 30)

= (x - 30)(x + 6) 

Solution 15

z2 - 32z - 105

= z2 - 35z + 3z - 105

= z (z - 35) + 3 (z - 35)

= (z - 35) (z + 3)

Solution 16

x2 - 11x - 80

= x2 - 16x + 5x - 80

= x (x - 16) + 5 (x - 16)

= (x - 16) (x + 5).

Solution 17

6 - x - x2

= 6 + 2x - 3x - x2

= 2(3 + x) - x (3 + x)

= (3 + x) (2 - x).

Solution 18

  

Solution 19

40 + 3x - x2

= 40 + 8x - 5x - x2

= 8 (5 + x) -x (5 + x)

= (5 + x) (8 - x).

Solution 20

x2 - 26x + 133

= x2 - 19x - 7x + 133

= x(x - 19) - 7(x - 19)

= (x - 19)(x - 7) 

Solution 21

  

Solution 22

  

Solution 23

  

Solution 24

  

Solution 25

x2 - x - 156

= x2 - 13x + 12x - 156

= x (x - 13) + 12 (x - 13)

= (x - 13) (x + 12).

Solution 27

9x2 + 18x + 8

= 9x2 + 12x + 6x + 8

= 3x (3x+ 4) +2 (3x + 4)

= (3x + 4) (3x + 2).

Solution 28

6x2 + 17x + 12

= 6x2 + 9x + 8x + 12

= 3x (2x + 3) + 4(2x + 3)

= (2x + 3) (3x + 4).

Solution 29

18x2 + 3x - 10

= 18x2 - 12x + 15x - 10

= 6x (3x - 2) + 5 (3x - 2)

= (6x + 5) (3x - 2).

Solution 30

2x2 + 11x - 21

= 2x2 + 14x - 3x - 21

= 2x (x + 7) - 3 (x + 7)

= (x + 7) (2x - 3).

Solution 31

15x2 + 2x - 8

= 15x2 - 10x + 12x - 8

= 5x (3x - 2) + 4 (3x - 2)

= (3x - 2) (5x + 4).

Solution 32

21x2 + 5x - 6

= 21x2 + 14x - 9x - 6

= 7x(3x + 2) - 3(3x + 2)

= (3x + 2)(7x - 3) 

Solution 33

24x2 - 41x + 12

= 24x2 - 32x - 9x + 12

= 8x (3x - 4) - 3 (3x - 4)

= (3x - 4) (8x - 3).

Solution 34

3x2 - 14x + 8

= 3x2 - 12x - 2x +8

= 3x (x - 4) - 2(x - 4)

= (x - 4) (3x - 2).

Solution 35

2x2 + 3x - 90

= 2x2 - 12x + 15x - 90

= 2x (x - 6) + 15 (x - 6)

= (x - 6) (2x + 15).

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

  

Solution 42

Solution 43

Solution 44

15x2 - x - 28

= 15x2 + 20x - 21x - 28

= 5x (3x + 4) - 7 (3x + 4)

= (3x + 4) (5x - 7).

Solution 45

6x2 - 5x - 21

= 6x2 + 9x - 14x - 21

= 3x (2x + 3) - 7 (2x + 3)

= (3x - 7) (2x + 3).

Solution 46

2x2 - 7x - 15

= 2x2 - 10x + 3x - 15

= 2x (x - 5) + 3 (x - 5)

= (x - 5) (2x + 3).

Solution 47

5x2 - 16x - 21

= 5x2 + 5x - 21x - 21

= 5x (x + 1) -21 (x + 1)

= (x + 1) (5x - 21).

Solution 48

6x2 - 11x - 35

= 6x2 - 21x + 10x - 35

= 3x(2x - 7) + 5(2x - 7)

= (2x - 7)(3x + 5) 

Solution 49

9x2 - 3x - 20

= 9x2 - 15x + 12x - 20

= 3x(3x - 5) + 4(3x - 5)

= (3x - 5)(3x + 4) 

Solution 50

10x2 - 9x - 7

= 10x2 + 5x - 14x - 7

= 5x (2x + 1) - 7 (2x+ 1)

= (2x + 1) (5x - 7).

Solution 51

x2 - 2x +

equals left parenthesis 4 straight x minus 1 right parenthesis open parentheses straight x over 4 minus 7 over 16 close parentheses

Solution 52

Solution 53

  

Solution 54

  

Solution 55

  

Solution 56

  

Solution 57

  

Solution 58

  

Solution 59

Let x + y = z

Then, 2 (x + y)2 - 9 (x + y) - 5

Now, replacing z by (x + y), we get

Solution 60

Let 2a - b = c

Then, 9 (2a - b)2 - 4 (2a - b) -13

Now, replacing c by (2a - b) , we get

9 (2a - b)2 - 4 (2a - b) - 13

Solution 61

Let x - 2y = z

Then, 7 (x - 2y)2 - 25 (x - 2y) + 12

Now replace z by (x - 2y), we get

7 (x - 2y)2 - 25 (x - 2y) + 12

Solution 62

  

Solution 63

  

Solution 64

Given equation: (a + 2b)2 + 101(a + 2b) + 100

Let (a + 2b) = x

Then, we have

x2 + 101x + 100

= x2 + 100x + x + 100

= x(x + 100) + 1(x + 100)

= (x + 100)(x + 1)

= (a + 2b + 100)(a + 2b + 1) 

Solution 65

Let x2 = y

Then, 4x4 + 7x2 - 2

Now replacing y by x2, we get

Solution 66

{(999)2 - 1}

= {(999)2 - (1)2}

= {(999 - 1)(999 + 1)}

= 998 × 1000

= 998000 

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials Page/Excercise 3D

Solution 1(i)

We know:

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(i) (a + 2b + 5c)2

= (a)2 + (2b)2 + (5c)2 + 2(a) (2b) + 2 (2b) (5c) + 2(5c) (a)

= a2 + 4b2 + 25c2 + 4ab + 20bc + 10ac

Solution 1(ii)

We know:

(2a - b + c)2

= (2a)2 + (-b)2 + (c)2 + 2 (2a) (-b) + 2(-b) (c) + 2 (c) (2a)

= 4a2 + b2 + c2 - 4ab - 2bc + 4ac.

Solution 1(iii)

We know:

(a - 2b - 3c)2

= (a)2 + (-2b)2 + (-3c)2+ 2(a) (-2b) + 2(-2b) (-3c) + 2 (-3c) (a)

= a2 + 4b2 + 9c2 - 4ab + 12bc - 6ac.

Solution 2(i)

We know:

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(2a - 5b - 7c)2

= (2a)2 + (-5b)2 + (-7c)2 + 2 (2a) (-5b) + 2 (-5b) (-7c) + 2 (-7c) (2a)

= 4a2 + 25b2 + 49c2 - 20ab + 70bc - 28ac.

Solution 2(ii)

(-3a + 4b - 5c)2

= (-3a)2 + (4b)2 + (-5c)2 + 2 (-3a) (4b) + 2 (4b) (-5c) + 2 (-5c) (-3a)

= 9a2 + 16b2 + 25c2 - 24ab - 40bc + 30ac.

Solution 2(iii)

=

Solution 3

4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz

= (2x)2 + (3y)2 + (-4z)2 + 2 (2x) (3y) + 2(3y) (-4z) + 2 (-4z) (2x)

= (2x + 3y - 4z)2

Solution 4

9x2 + 16y2 + 4z2 - 24xy + 16yz - 12xz

= (-3x)2 + (4y)2 + (2z)2 + 2 (-3x) (4y) + 2 (4y) (2z) + 2 (2z) (-3x)

= (-3x + 4y + 2z)2.

Solution 5

25x2 + 4y2 + 9z2 - 20xy - 12yz + 30xz

= (5x)2 + (-2y)2 + (3z)2 + 2(5x) (-2y) + 2(-2y) (3z) + 2(3z) (5x)

= (5x - 2y + 3z)2

Solution 6

16x2 + 4y2 + 9z2 - 16xy - 12yz + 24xz

= (4x)2 + (-2y)2 + (3z)2 + 2(4x)(-2y) + 2(-2y)(3z) + 2(4x)(3z)

= [4x + (-2y) + 3z]2

= (4x - 2y + 3z)2

= (4x - 2y + 3z)(4x - 2y + 3z) 

Solution 7(ii)

(995)2

= (1000 - 5)2

= (1000)2 + (5)2 - 2(1000)(5)

= 1000000 + 25 - 10000

= 990025 

Solution 7(iii)

(107)2

= (100 + 7)2

= (100)2 + (7)2 + 2(100)(7)

= 10000 + 49 + 1400

= 11449 

Solution 7(i)

(99)2

= (100 - 1)2

= (100)2 - 2(100) (1) + (1)2

= 10000 - 200 + 1

= 9801.

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials Page/Excercise 3A

Solution 1

9x2 + 12xy = 3x (3x + 4y)

Solution 2

18x2y - 24xyz = 6xy (3x - 4z)

Solution 3

27a3b3 - 45a4b2 = 9a3b2 (3b - 5a)

Solution 4

2a (x + y) - 3b (x + y) = (x + y) (2a - 3b)

Solution 5

2x (p2 + q2) + 4y (p2 + q2)

= (2x + 4y) (p2 + q2)

= 2(x+ 2y) (p2 + q2)

Solution 6

x (a - 5) + y (5 - a)

= x (a - 5) + y (-1) (a - 5)

= (x - y) (a - 5)

Solution 7

4 (a + b) - 6 (a + b)2

= (a + b) [4 - 6 (a + b)]

= 2 (a + b) (2 - 3a - 3b)

= 2 (a + b) (2 - 3a - 3b)

Solution 8

8 (3a - 2b)2 - 10 (3a - 2b)

= (3a - 2b) [8(3a - 2b) - 10]

= (3a - 2b) 2[4 (3a - 2b) - 5]

= 2 (3a - 2b) (12 a - 8b - 5)

Solution 9

x (x + y)3 - 3x2y (x + y)

= x (x + y) [(x + y)2 - 3xy]

= x (x + y) (x2 + y2 + 2xy - 3xy)

= x (x + y) (x2 + y2 - xy)

Solution 10

x3 + 2x2 + 5x + 10

= x2 (x + 2) + 5 (x + 2)

= (x2 + 5) (x + 2)

Solution 11

x2 + xy - 2xz - 2yz

= x (x + y) - 2z (x + y)

= (x+ y) (x - 2z)

Solution 12

a3b - a2b + 5ab - 5b

= a2b (a - 1) + 5b (a - 1)

= (a - 1) (a2b + 5b)

= (a - 1) b (a2 + 5)

= b (a - 1) (a2 + 5)

Solution 13

8 - 4a - 2a3 + a4

= 4(2 - a) - a3 (2 - a)

= (2 - a) (4 - a3)

Solution 14

x3 - 2x2y + 3xy2 - 6y3

= x2 (x - 2y) + 3y2 (x - 2y)

= (x - 2y) (x2 + 3y2)

Solution 15

px + pq - 5q - 5x

= p(x + q) - 5 (q + x)

= (x + q) (p - 5)

Solution 16

x2 - xy + y - x

= x (x - y) - 1 (x - y)

= (x - y) (x - 1)

Solution 17

(3a - 1)2 - 6a + 2

= (3a - 1)2 - 2 (3a - 1)

= (3a - 1) [(3a - 1) - 2]

= (3a - 1) (3a - 3)

= 3(3a - 1) (a - 1)

Solution 18

(2x - 3)2 - 8x + 12

= (2x - 3)2 - 4 (2x - 3)

= (2x - 3) (2x - 3 - 4)

= (2x - 3) (2x - 7)

Solution 19

a3 + a - 3a2 - 3

= a(a2 + 1) - 3 (a2 + 1)

= (a - 3) (a2 + 1)

Solution 20

3ax - 6ay - 8by + 4bx

= 3a (x - 2y) + 4b (x - 2y)

= (x - 2y) (3a + 4b)

Solution 21

abx2 + a2x + b2x +ab

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)

Solution 22

x3 - x2 + ax + x - a - 1

= x3 - x2 + ax - a + x - 1

= x2 (x - 1) + a (x - 1) + 1 (x - 1)

= (x - 1) (x2 + a + 1)

Solution 23

2x + 4y - 8xy - 1

= 2x - 1 - 8xy + 4y

= (2x - 1) - 4y (2x - 1)

= (2x - 1) (1 - 4y)

Solution 24

ab (x2 + y2) - xy (a2 + b2)

= abx2 + aby2 - a2xy - b2xy

= abx2 - a2xy + aby2 - b2xy

= ax (bx - ay) + by(ay - bx)

= (bx - ay) (ax - by)

Solution 25

a2 + ab (b + 1) + b3

= a2 + ab2 + ab + b3

= a2 + ab + ab2 + b3

= a (a + b) + b2 (a + b)

= (a + b) (a + b2)

Solution 26

a3 + ab (1 - 2a) - 2b2

= a3 + ab - 2a2b - 2b2

= a (a2 + b) - 2b (a2 + b)

= (a2 + b) (a - 2b)

Solution 27

2a2 + bc - 2ab - ac

= 2a2 - 2ab - ac + bc

= 2a (a - b) - c (a - b)

= (a - b) (2a - c)

Solution 28

(ax + by)2 + (bx - ay)2

= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 - 2abxy

= a2x2 + b2y2 + b2x2 + a2y2

= a2x2 + b2x2 + b2y2 + a2y2

= x2 (a2 + b2) + y2(a2 + b2)

= (a2 + b2) (x2 + y2)

Solution 29

a (a + b - c) - bc

= a2 + ab - ac - bc

= a(a + b) - c (a + b)

= (a - c) (a + b)

Solution 30

a(a - 2b - c) + 2bc

= a2 - 2ab - ac + 2bc

= a (a - 2b) - c (a - 2b)

= (a - 2b) (a - c)

Solution 31

a2x2 + (ax2 + 1)x + a

= a2x2 + ax3 + x + a

= ax2 (a + x) + 1 (x + a)

= (ax2 + 1) (a + x)

Solution 32

ab (x2 + 1) + x (a2 + b2)

= abx2 + ab + a2x + b2x

= abx2 + a2x + ab + b2x

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)

Solution 33

x2 - (a + b) x + ab

= x2 - ax - bx + ab

= x (x - a) - b(x - a)

= (x - a) (x - b)

Solution 34

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials Page/Excercise 3F

Solution 1

x3 + 27

= x3 + 33

= (x + 3) (x2 - 3x + 9)

Solution 2

27a3 + 64b3

= (3a)3 + (4b)3

= (3a + 4b)[(3a)2 - (3a)(4b) + (4b)2]

= (3a + 4b)(9a2 - 12ab + 16b2) 

Solution 3

125a3 +

We know that

Let us rewrite



Solution 4

216x3 +

We know that

Let us rewrite



Solution 5

16x 4 + 54x

= 2x (8x 3 + 27)

= 2x [(2x)3 + (3)3]

= 2x (2x + 3) [(2x)2 - 2x3 + 32]   

=2x(2x+3)(4x2 -6x +9)


Solution 6

7a3 + 56b3

= 7(a3 + 8b3)

= 7 [(a)3 + (2b)3]

= 7 (a + 2b) [a2 - a 2b + (2b)2]   

= 7 (a + 2b) (a2 - 2ab + 4b2).


Solution 7

x5 + x2

= x2(x3 + 1)

= x2 (x + 1) [(x)2 - x 1 + (1)2]    

= x2 (x + 1) (x2 - x + 1).


Solution 8

a3 + 0.008

= (a)3 + (0.2)3

= (a + 0.2) [(a)2 - a 0.2 + (0.2)2]    

= (a + 0.2) (a2 - 0.2a + 0.04).


Solution 9

1 - 27x3

= (1)3 - (3x)3

= (1 - 3x) [(1)2 + 1 3x + (3x)2]      

= (1 - 3x) (1 + 3x + 9x2).


Solution 10

64a3 - 343

= (4a)3 - (7)3

= (4a - 7)[(4a)2 + (4a)(7) + (7)2]

= (4a - 7)(16a2 + 28a + 49) 

Solution 11

x3 - 512

= (x)3 - (8)3

= (x - 8) [(x)2 + x 8 + (8)2]      

= (x - 8) (x2 + 8x + 64).


Solution 12

a3 - 0.064

= (a)3 - (0.4)3

= (a - 0.4) [(a)2 + a 0.4 + (0.4)2]    

= (a - 0.4) (a2 + 0.4 a + 0.16).


Solution 13

We know that

Let us rewrite


Solution 14

  

Solution 15

x - 8xy3

= x (1 - 8y3)

= x [(1)3 - (2y)3]

= x (1 - 2y) [(1)2 + 1 2y + (2y)2]   

= x (1 - 2y) (1 + 2y + 4y2).


Solution 16

32x4 - 500x

= 4x (8x3 - 125)

= 4x [(2x)3 - (5)3]

= 4x [(2x - 5) [(2x)2 + 2x 5 + (5)2]     

= 4x (2x - 5) (4x2 + 10x + 25).


Solution 17

3a7b - 81a4b4

= 3a4b (a3 - 27b3)

= 3a4b [(a)3 - (3b)3]

= 3a4b (a - 3b) [(a)2 + a 3b + (3b)2]     

= 3a4b (a - 3b) (a2 + 3ab + 9b2).


Solution 18

x4y4 - xy

= xy(x3y3 - 1)

= xy[(xy)3 - (1)3]

= xy{(xy - 1)[(xy)2 + (xy)(1) + (1)2]}

= xy(xy - 1)(x2y2 + xy + 1) 

Solution 19

8x2y3 - x5

= x2 (8y3 - x3)

= x2 [(2y)3 - x3]

= x2 [(2y - x)[(2y)2 + (2y)(x) + x2]

= x2 (2y - x)(4y2 + 2xy + x2) 

Solution 20

1029 - 3x3

= 3(343 - x3)

= 3[(7)3 - x3]

= 3[(7 - x)(72 + 7x + x2)]

= 3(7 - x)(49 + 7x + x2) 

Solution 21

x6 - 729

= (x2)3 - (9)3

= (x2 - 9) [(x2)2 + x2 9 + (9)2]     

= (x2 - 9) (x4 + 9x2 + 81)

= (x + 3) (x - 3) [(x2 + 9)2 - (3x)2]

= (x + 3) (x - 3) (x2 + 3x + 9) (x2 - 3x + 9).


Solution 22

x9 - y9

= (x3)3 - (y3)3 

= [(x3 - y3)][(x3)2 + x3y3 + (y3)2]

= [(x - y)(x2 + xy + y2)(x6 + x3y3 + y6)  

Solution 23

We know that,

Therefore,

(a + b)3 - (a - b)3

= [a + b - (a - b)] [ (a + b)2 + (a + b) (a - b) + (a - b)2

= (a + b - a + b) [ a2 + b2 + 2ab + a2 - b2 + a2 + b2 - 2ab]

= 2b (3a2 + b2).


Solution 24

8a3 - b3 - 4ax + 2bx

= 8a3 - b3 - 2x (2a - b)

= (2a)3 - (b)3 - 2x (2a - b)

= (2a - b) [(2a)2 + 2a b + (b)2] - 2x (2a - b)   

= (2a - b) (4a2 + 2ab + b2) - 2x (2a - b)

= (2a - b) (4a2 + 2ab + b2 - 2x).


Solution 25

begin mathsize 14px style straight a cubed plus 3 straight a squared straight b plus 3 ab squared plus straight b cubed minus 8
equals left parenthesis straight a plus straight b right parenthesis cubed minus 2 cubed
equals left square bracket left parenthesis straight a plus straight b right parenthesis minus 2 right square bracket space left square bracket left parenthesis straight a plus straight b right parenthesis squared plus left parenthesis straight a plus straight b right parenthesis 2 plus 2 squared right square bracket
equals left parenthesis straight a plus straight b minus 2 right parenthesis left square bracket left parenthesis straight a plus straight b right parenthesis squared plus 2 left parenthesis straight a plus straight b right parenthesis plus 4 right square bracket end style

Solution 26

We know that

Solution 27

2a3 + 16b3 - 5a - 10b

= 2 (a3 + 8b3) - 5 (a + 2b)

= 2 [(a)3 + (2b)3] - 5 (a + 2b)

= 2 (a + 2b) [(a)2 - a 2b + (2b)2 ] - 5 (a + 2b)    

= (a + 2b) [2(a2 - 2ab + 4b2) - 5]


Solution 28

a6 + b6

= (a2)3 + (b2)3 

= (a2 + b2)[(a2)2 - (a2b2) + (b2)2]

= (a2 + b2)(a4 - a2b2 + b4) 

Solution 29

a12 - b12

= (a6)2 - (b6)2 

= (a6 - b6)(a6 + b6)

= [(a3)2 - (b3)2][(a2)3 + (b2)3]

= (a3 - b3)(a3 + b3)[(a2 + b2)(a4 - a2b2 + b4)] 

= (a - b)(a2 + ab + b2)(a + b)(a2 - ab + b2)(a2 + b2)(a4 - a2b2 + b4)

= (a - b)(a + b)(a2 + b2)(a2 + ab + b2)(a2 - ab + b2)(a4 - a2b2 + b4)

Solution 30

Given equation is x6 - 7x3 - 8.

Putting x3 = y, we get

y2 - 7y - 8

= y2 - 8y + y - 8

= y(y - 8) + 1(y - 8)

= (y - 8)(y + 1)

= (x3 - 8)(x3 + 1)

= (x3 - 23)(x3 + 13)

= (x - 2)(x2 + 2x + 4)(x + 1)(x2 - x + 1)

= (x - 2)(x + 1)(x2 + 2x + 4)(x2 - x + 1) 

Solution 31

x3 - 3x2 + 3x + 7

= x3 - 3x2 + 3x - 1 + 8

= (x3 - 3x2 + 3x - 1) + 8

= (x - 1)3 + 23

= (x - 1 + 2)[(x - 1)2 - (x - 1)(2) + 22]

= (x + 1)(x2 - 2x + 1 - 2x + 2 + 4)

= (x + 1)(x2 - 4x + 7) 

Solution 32

(x + 1)3 + (x - 1)3

= (x + 1 + x - 1)[(x + 1)2 - (x + 1)(x - 1) + (x - 1)2]

= 2x(x2 + 2x + 1 - x2 + 1 + x2 - 2x + 1)

= 2x(x2 + 3)  

Solution 33

(2a + 1)3 + (a - 1)3

= (2a + 1 + a - 1)[(2a + 1)2 - (2a + 1)(a - 1) + (a - 1)2]

= (3a)[4a2 + 4a + 1 - 2a2 + 2a - a + 1 + a2 - 2a + 1]

= 3a(3a2 + 3a + 3)

= 9a(a2 + a + 1)  

Solution 34

8(x + y)3 - 27(x - y)3

= [23 (x + y)3] - [33 (x - y)3]

= [2(x + y) - 3(x - y)]{[2(x + y)]2 + 2(x + y)3(x - y) + [3(x - y)]2}

= (2x + 2y - 3x + 3y){[4(x2 + y2 + 2xy)] + 6(x2 - y2) + [9(x2 + y2 - 2xy]}

= (-x + 5y){4x2 + 4y2 + 8xy + 6x2 - 6y2 + 9x2 + 9y2 - 18xy}

= (-x + 5y)(19x2 + 7y2 - 10xy) 

Solution 35

(x + 2)3 + (x - 2)3

= [(x + 2) + (x - 2)][(x + 2)2 - (x + 2)(x - 2) + (x - 2)2]

=(2x)(x2 + 4x + 4 - x2 + 4 + x2 - 4x + 4)

= 2x(x2 + 12) 

Solution 36

(x + 2)3 - (x - 2)3

= [(x + 2) - (x - 2)][(x + 2)2 + (x + 2)(x - 2) + (x - 2)2]

= 4(x2 + 4x + 4 + x2 - 4 + x2 - 4x + 4)

= 4(3x2 + 4) 

Solution 37

  

Solution 38

  

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials Page/Excercise 3B

Solution 1

9x2 - 16y2

= (3x)2 - (4y)2

= (3x + 4y)(3x - 4y) 

Solution 2

  

Solution 3

81 - 16x2

= (9)2 - (4x)2

= (9 - 4x)(9 + 4x) 

Solution 4

5 - 20x2

= 5(1 - 4x2)

= 5[(1)2 - (2x)2]

= 5[(1 - 2x)(1 + 2x)]

= 5(1 - 2x)(1 + 2x) 

Solution 5

2x4 - 32

= 2(x4 - 16)

= 2[(x2)2 - (4)2]

= 2[(x2 - 4)(x2 + 4)]

= 2[(x2 - 22)(x2 + 4)]

= 2[(x - 2)(x + 2)(x2 + 4)]

= 2(x - 2)(x + 2)(x2 + 4) 

Solution 6

3a3b - 243ab3

= 3ab (a2 - 81 b2)

= 3ab [(a)2 - (9b)2]

= 3ab (a + 9b) (a - 9b)

Solution 7

3x3 - 48x

= 3x (x2 - 16)

= 3x [(x)2 - (4)2]

= 3x (x + 4) (x - 4)

Solution 8

27a2 - 48b2

= 3 (9a2 - 16b2)

= 3 [(3a)2 - (4b)2]

= 3(3a + 4b) (3a - 4b)

Solution 9

x - 64x3

= x (1 - 64x2)

= x[(1)2 - (8x)2]

= x (1 + 8x) (1 - 8x)

Solution 10

8ab2 - 18a3

= 2a (4b2 - 9a2)

= 2a [(2b)2 - (3a)2]

= 2a (2b + 3a) (2b - 3a)

Solution 11

150 - 6x2

= 6 (25 - x2)

= 6 (52 - x2)

= 6 (5 + x) (5 - x)

Solution 12

2 - 50x2

= 2 (1 - 25x2)

= 2 [(1)2 - (5x)2]

= 2 (1 + 5x) (1 - 5x)

Solution 13

20x2 - 45

= 5(4x2 - 9)

= 5 [(2x)2 - (3)2]

= 5 (2x + 3) (2x - 3)

Solution 14

(3a + 5b)2 - 4c2

= (3a + 5b)2 - (2c)2

= (3a + 5b - 2c)(3a + 5b + 2c) 

Solution 15

a2 - b2 - a - b

= a2 - b2 - (a + b)

= (a - b)(a + b) - (a + b)

= (a + b)(a - b - 1) 

Solution 16

4a2 - 9b2 - 2a - 3b

= (2a)2 - (3b)2 - (2a + 3b)

= (2a - 3b)(2a + 3b) - (2a + 3b)

= (2a + 3b)(2a - 3b - 1) 

Solution 17

a2 - b2 + 2bc - c2

= a2 - (b2 - 2bc + c2)

= a2 - (b - c)2

= [a - (b - c)][a + (b - c)]

= (a - b + c)(a + b - c) 

Solution 18

4a2 - 4b2 + 4a + 1

= (4a2 + 4a + 1) - 4b2

= [(2a)2 + 2 × 2a × 1 + (1)2] - (2b)2

= (2a + 1)2 - (2b)2

= (2a + 1 - 2b)(2a + 1 + 2b)

= (2a - 2b + 1)(2a + 2b + 1) 

Solution 19

a2 + 2ab + b2 - 9c2

= (a + b)2 - (3c)2

= (a + b + 3c) (a + b - 3c)

Solution 20

108a2 - 3(b - c)2

= 3 [(36a2 - (b -c)2]

= 3 [(6a)2 - (b - c)2]

= 3 (6a + b - c) (6a - b + c)

Solution 21

(a + b)3 - a - b

= (a + b)3 - (a + b)

= (a + b) [(a + b)2 - 12]

= (a + b) (a + b + 1) (a + b - 1)

Solution 22

x2 + y2 - z2 - 2xy

= (x2 + y2 - 2xy) - z2

= (x - y)2 - z2

= (x - y - z)(x - y + z) 

Solution 23

x2 + 2xy + y2 - a2 + 2ab - b2

= (x2 + 2xy + y2) - (a2 - 2ab + b2)

= (x + y)2 - (a - b)2

= [(x + y) - (a - b)][(x + y) + (a - b)]

= (x + y - a + b)(x + y + a - b) 

Solution 24

25x2 - 10x + 1 - 36y2

= (25x2 - 10x + 1) - 36y2

= [(5x)2 - 2(5x)(1) + (1)2] - (6y)2

= (5x - 1)2 - (6y)2

= (5x - 1 - 6y)(5x - 1 + 6y) 

Solution 25

a - b - a2 + b2

= (a - b) - (a2 - b2)

= (a - b) - (a - b) (a + b)

= (a - b) (1 - a - b)

Solution 26

a2 - 4ac + 4c2 - b2

= a2 - 4ac + 4c2 - b2

= a2 - 2 a 2c + (2c)2 - b2

= (a - 2c)2 - b2

= (a - 2c + b) (a - 2c - b)

Solution 27

9 - a2 + 2ab - b2

= 9 - (a2 - 2ab + b2)

= 32 - (a - b)2

= (3 + a - b) (3 - a + b)

Solution 28

x3 - 5x2 - x + 5

= x2 (x - 5) - 1 (x - 5)

= (x - 5) (x2 - 1)

= (x - 5) (x + 1) (x - 1)

Solution 29

1 + 2ab - (a2 + b2)

= 1 - (a2 + b2 - 2ab)

= (1)2 - (a - b)2

= [1 - (a - b)][1 + (a - b)]

= (1 - a + b)(1 + a - b) 

Solution 30

9a2 + 6a + 1 - 36b2

= (9a2 + 6a + 1) - 36b2

= [(3a)2 + 2(3a)(1) + (1)2] - (6b)2

= (3a + 1)2 - (6b)2

= (3a + 1 - 6b)(3a + 1 + 6b) 

Solution 31

x2 - y2 + 6y - 9

= x2 - (y2 - 6y + 9)

= x2 - (y2 - 2 y 3 + 32)

= x2 - (y - 3)2

= [x + (y - 3)] [x - (y - 3)]

= (x + y - 3) (x - y + 3)

Solution 32

4x2 - 9y2 - 2x - 3y

= (2x)2 - (3y)2 - (2x + 3y)

= (2x + 3y) (2x - 3y) - (2x + 3y)

= (2x + 3y) (2x - 3y - 1)

Solution 33

9a2 + 3a - 8b - 64b2

= 9a2 - 64b2 + 3a - 8b

= (3a)2 - (8b)2 + (3a - 8b)

= (3a + 8b) (3a - 8b) + (3a - 8b)

= (3a - 8b) (3a + 8b + 1)

Solution 34

  

Solution 35

  

Solution 36

  

Solution 37

x8 - 1

= (x4)2 - (1)2

= (x4 - 1)(x4 + 1)

= [(x2)2 - (1)2)(x4 + 1)

= (x2 - 1)(x2 + 1)(x4 + 1)

= (x - 1)(x + 1)(x2 + 1)[(x2)2 + (1)2 + 2x2 - 2x2

= (x - 1)(x + 1)(x2 + 1)[(x2)2 + (1)2 + 2x2) - 2x2

Solution 38

16x4 - 1

= (4x2)2 - (1)2

= (4x2 - 1)(4x2 + 1)

= [(2x)2 - (1)2](4x2 + 1)

= (2x - 1)(2x + 1)(4x2 + 1) 

Solution 39

81x4 - y4

= (9x2)2 - (y2)2

= (9x2 - y2)(9x2 + y2)

= [(3x)2 - y2](9x2 + y2)

= (3x - y)(3x + y)(9x2 + y2) 

Solution 40

x4 - 625

= (x2)2 - (25)2

= (x2 + 25) (x2 - 25)

= (x2 + 25) (x2 - 52)

= (x2 + 25) (x + 5) (x - 5)

R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 3 - Factorisation of Polynomials Page/Excercise 3G

Solution 1

(x + y - z) (x2 + y2 + z2 - xy + yz + zx)

= [x + y + (-z)] [(x)2 + (y)2 + (-z)2 - (x) (y) - (y) (-z) - (-z) (x)]

= x3 + y3 - z3 + 3xyz.

Solution 2

(x - y - z)(x2 + y2 + z2 + xy - yz + xz)

= x3 + xy2 + xz2 + x2y  - xyz + x2z - x2y - y3 - yz2 - xy2 + y2z - xyz - x2z - y2z - z3 - xyz + yz2 - xz2

= x3 - y3 - z3 - xyz - xyz - xyz

= x3 - y3 - z3 - 3xyz 

Solution 3

(x - 2y + 3) (x2 + 4y2 + 2xy - 3x + 6y + 9)

= [x + (-2y) + 3] [(x)2 + (-2y)2 + (3) - (x) (-2y) - (-2y) (3) - (3) (x)]

= (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

= a3 + b3 + c3 - 3abc

Where, x = a, (-2y) = b and 3 = c

(x - 2y + 3) (x2 + 4y2 + 2xy - 3x + 6y + 9)

= (x)3 + (-2y)3 + (3)2 - 3 (x) (-2y) (3)

= x3 - 8y3 + 27 + 18xy.

Solution 4

(3x - 5y + 4)(9x2 + 25y2 + 15xy + 20y - 12x + 16)

= (3x + (-5y) + 4)[(3x)2 + (-5y)2 + (4)2 - (3x)(-5y) - (-5y)(4) - (3x)(4)]

= (3x)3 + (-5y)3 + (4)3 - 3(3x)(-5y)(4)

= 27x3 - 125y3 + 64 + 180xy 

Solution 5

125a3 + b3 + 64c3 - 60abc

= (5a)3 + (b)3 + (4c)3 - 3 (5a) (b) (4c)

= (5a + b + 4c) [(5a)2 + b2 + (4c)2 - (5a) (b) - (b) (4c) - (5a) (4c)]

[ a3 + b3 + c3 - 3abc = (a+ b + c) (a2 + b2 + c2 - ab - bc - ca)]

= (5a + b + 4c) (25a2 + b2 + 16c2 - 5ab - 4bc - 20ac).

Solution 6

a3 + 8b3 + 64c3 - 24abc

= (a)3 + (2b)3 + (4c)3 - 3 a 2b 4c

= (a + 2b + 4c) [a2 + 4b2 + 16c2 - 2ab - 8bc - 4ca).

Solution 7

1 + b3 + 8c3 - 6bc

= 1 + (b)3 + (2c)3 - 3 (b) (2c)

= (1 + b + 2c) [1 + b2 + (2c)2 - b - b 2c - 2c]

= (1 + b + 2c) (1 + b2 + 4c2 - b - 2bc - 2c).

Solution 8

216 + 27b3 + 8c3 - 108bc

= (6)3 + (3b)3 + (2c)2 - 3 6 3b 2c

= (6 + 3b + 2c) [(6)2 + (3b)2 + (2c)2 - 6 3b - 3b 2c - 2c 6]

= (6 + 3b + 2c) (36 + 9b2 + 4c2 - 18b - 6bc - 12c).

Solution 9

27a3 - b3 + 8c3 + 18abc

= (3a)3 + (-b)3 + (2c)3 + 3(3a) (-b) (2c)

= [3a + (-b) + 2c] [(3a)2 + (-b)2 + (2c)2 - 3a (-b) - (-b) (2c) - (2c) (3a)]

= (3a - b + 2c) (9a2 + b2 + 4c2 + 3ab + 2bc - 6ca).

Solution 10

8a3 + 125b3 - 64c3 + 120abc

= (2a)3 + (5b)3 + (-4c)3 - 3 (2a) (5b) (-4c)

= (2a + 5b - 4c) [(2a)2 + (5b)2 + (-4c)2 - (2a) (5b) - (5b) (-4c) - (-4c) (2a)]

= (2a + 5b - 4c) (4a2 + 25b2 + 16c2 - 10ab + 20bc + 8ca).

Solution 11

8 - 27b3 - 343c3 - 126bc

= (2)3 + (-3b)3 + (-7c)3 - 3(2) (-3b) (-7c)

= (2 - 3b - 7c) [(2)2 + (-3b)2 + (-7c)2 - (2) (-3b) - (-3b) (-7c) - (-7c) (2)]

= (2 - 3b - 7c) (4 + 9b2 + 49c2 + 6b - 21bc + 14c).

Solution 12

125 - 8x3 - 27y3 - 90xy

= (5)3 + (-2x)3 + (-3y)3 - 3 (5) (-2x) (-3y)

= (5 - 2x - 3y) [(5)2 + (-2x)2 + (-3y)2 - (5) (-2x) - (-2x) (-3y) - (-3y) (5)]

= (5 - 2x - 3y) (25 + 4x2 + 9y2 + 10x - 6xy + 15y).

Solution 13

Solution 14

27x3 - y3 - z3 - 9xyz

= (3x)3 - y3 - z3 - 3(3x)(y)(z)

= (3x)3 + (-y)3 + (-z)3 - 3(3x)(-y)(-z)

= [3x + (-y) + (-z)][(3x)2 + (-y)2 + (-z)2 - (3x)(-y) - (-y)(-z) - (3x)(-z)]

= (3x - y - z)(9x2 + y2 + z2 + 3xy - yz + 3zx) 

Solution 15

  

Solution 16

  

Solution 17

Putting (a - b) = x, (b - c) = y and (c - a) = z, we get,

(a - b)3 + (b - c)3 + (c - a)3

= x3 + y3 + z3, where (x + y + z) = (a - b) + (b - c) + (c - a) = 0

= 3xyz [ (x + y + z) = 0 (x3 + y3 + z3) = 3xyz]

= 3(a - b) (b - c) (c - a).

Solution 18

Given equation is (a - 3b)3 + (3b - c)3 + (c - a)3

Now,

(a - 3b) + (3b - c) + (c - a)

= a - a - 3b + 3b - c + c

= 0

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz

Hence,

(a - 3b)3 + (3b - c)3 + (c - a)3

= 3(a - 3b)(3b - c)(c - a)

Solution 19

We have:

(3a - 2b) + (2b - 5c) + (5c - 3a) = 0

So, (3a - 2b)3 + (2b - 5c)3 + (5c - 3a)3

= 3(3a - 2b) (2b - 5c) (5c - 3a).

Solution 20

Given equation is (5a - 7b)3 + (7b - 9c)3 + (9c - 5a)3

Now,

(5a - 7b) + (7b - 9c) + (9c - 5a)

= 5a - 7b + 7b - 9c + 9c - 5a

= 0

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz

Hence,

(5a - 7b)3 + (7b - 9c)3 + (9c - 5a)3

= 3(5a - 7b)(7b - 9c)(9c - 5a) 

Solution 21

a3 (b - c)3 + b3 (c - a)3 + c3 (a - b)3

= [a (b - c)]3 + [b (c - a)]3 + [c (a - b)]3

Now, since, a (b - c) + b (c -a) + c (a - b)

= ab - ac + bc - ba + ca - bc = 0

So, a3 (b - c)3 + b3 (c - a)3 + c3 (a - b)3

= 3a (b - c) b (c - a) c (a - b)

= 3abc (a - b) (b - c) (c - a).

Solution 22(i)

Given equation is (-12)3 + 73 + 53

Now,

-12 + 7 + 5 = -12 + 12 = 0

We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz

Hence,

(-12)3 + 73 + 53

= 3(-12)(7)(5)

= -1260 

Solution 22(ii)

Given equation is (28)3 + (-15)3 + (-13)3

Now,

28 + (-15) + (-13) = 28 - 28 = 0

We know that if x + y + z = 0, then x3 +  y3 + z3 = 3xyz

Hence,

(28)3 + (-15)3 + (-13)3

= 3(28)(-15)(-13)

= 16380 

Solution 23

L.H.S. = (a + b + c)3 - a3 - b3 - c3

= [(a + b) + c]3 - a3 - b3 - c3

= (a + b)3 + c3 + 3(a + b)c × [(a + b) + c] - a3 - b3 - c3

= a3 + b3 + 3ab(a + b) + c3 + 3(a + b)c × [(a + b) + c] - a3 - b3 - c3

= 3ab(a + b) + 3(a + b)c × [(a + b) + c]

= 3(a + b)[ab + c(a + b) + c2]

= 3(a + b)[ab + ac + bc + c2]

= 3(a + b)[a(b + c) + c(b + c)]

= 3(a + b)[(b + c)(a + c)]

= 3(a + b)(b + c)(c + a)

= R.H.S.

Solution 24

  

Solution 25

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

= (a + b + c)[(a2 + b2 + c2) - (ab + bc + ca)] ….(i)

Now,

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

(9)2 = 35 + 2(ab + bc + ca)

81 = 35 + 2(ab + bc + ca)

2(ab + bc + ca) = 46

ab + bc + ca = 23

Substituting in (i), we get

a3 + b3 + c3 - 3abc = (9)[35 - 23] = 9 × 12 = 108 

TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 3 - Factorisation of Polynomials for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 3 - Factorisation of Polynomials now.

Text Book Solutions

CBSE IX - Mathematics

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